MHT CET · Maths · Application of Derivatives
Two positive numbers \(\mathrm{x}\) and \(\mathrm{y}\) such that \((x+y)=60\) and \(x y^3\) is maximum, then the numbers are respectively
- A 45,15
- B 30,30
- C 20,40
- D 40,20
Answer & Solution
Correct Answer
(A) 45,15
Step-by-step Solution
Detailed explanation
\(x+y=60 \Rightarrow x=60-y\)
Now let \(z=x y^3=(60-y) y^3=60 y^3-y^4\)
\(\Rightarrow \frac{d z}{d y}=180 y^2-4 y^3=4 y^2(45-y)\)
i.e., \(x y^3\) is maximum \(y=45\) and \(x=60-45=15\)
Now let \(z=x y^3=(60-y) y^3=60 y^3-y^4\)
\(\Rightarrow \frac{d z}{d y}=180 y^2-4 y^3=4 y^2(45-y)\)
i.e., \(x y^3\) is maximum \(y=45\) and \(x=60-45=15\)
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